Probability

Scenario

Your IPL fantasy team needs both Rohit Sharma and Jasprit Bumrah to perform in the same match. You want to know the combined chance.

Do you add or multiply individual probabilities when you need BOTH events to happen?

Mental model

Probability = favourable branches / all branches. Just count carefully.

It's a fraction: numerator is the ways the event you want can happen, denominator is all the equally likely outcomes. The hard part isn't the formula — it's counting both numbers correctly. That's why P&C is the prerequisite.

Formula

P(E) = (number of favourable outcomes) / (total outcomes). P(not E) = 1 - P(E). P(A or B) = P(A) + P(B) - P(A and B). P(A and B) = P(A) x P(B) [if independent]. nCr = n! / (r! x (n-r)!) for counting combinations within probability. Counting from groups: favourable = product of choices from each group; total = nCr from the whole pool.

'AND' events multiply probabilities. 'OR' events add then subtract the overlap. When drawing without replacement, total outcomes shrink with each draw. Use nCr to count the ways to pick r items from n when order doesn't matter.

Worked example

Problem: From a deck of 52 cards, 2 cards are drawn at random. What is the probability that both are Kings?

1. Set up: Drawing 2 from 52, order doesn't matter — use combinations.
2. Total outcomes: 52C2 = (52 x 51) / 2 = 1326.
3. Favourable outcomes (both Kings): there are 4 Kings in a deck. 4C2 = (4 x 3) / 2 = 6.
4. Apply: P(both Kings) = 6 / 1326 = 1/221.
5. Sanity-check: P(first King) = 4/52. P(second King | first King drawn) = 3/51. Product = 12/(52x51) = 12/2652 = 1/221. Matches.

Answer: 1/221.

Problem: In 4 coin tosses, what is the probability of getting at least 1 Head?

1. Set up: 'At least 1' complement trick: P(at least 1 Head) = 1 - P(no Heads at all).
2. Total outcomes: 2^4 = 16.
3. P(no Heads) = P(all Tails) = (1/2)^4 = 1/16.
4. Apply: P(at least 1 Head) = 1 - 1/16 = 15/16.
5. Sanity-check: Only 1 way to get all Tails (TTTT). So 15 of the 16 outcomes have at least one Head. Correct.

Answer: 15/16.

Shortcuts

• The 'At Least One' Trick: P(at least 1) = 1 - P(none). Always faster than summing P(exactly 1) + P(exactly 2) + ... (Probability of at least one Head in 4 tosses = 1 - (1/2)^4 = 1 - 1/16 = 15/16.)
• AND vs OR: Independent events AND = multiply. Mutually exclusive events OR = add. Mixed OR = add then subtract overlap. (P(Head AND 6 on dice) = 1/2 x 1/6 = 1/12.)

Common mistakes

• Adding independent probabilities instead of multiplying: P(Head AND roll 6) = 1/2 x 1/6 = 1/12, NOT 1/2 + 1/6. 'AND' = multiply for independent events.
• Forgetting without-replacement shrinks the denominator: Drawing two cards from a deck without replacing: first card is from 52, second is from 51. Don't use 52 both times.

Glossary

denominator

The bottom number in a fraction.

numerator

The top number in a fraction.

independent

Events that do not affect each other outcomes.

Recall questions

• P(at least 1 Head in 3 coin tosses) = ?
• If P(E) = 0.3, what is P(not E)?
• When drawing 2 balls from a bag of 10, how do you count total outcomes?

Questions & answers

A bag contains 4 red and 6 black balls. 2 balls are drawn at random. What is the probability that both are red?

2/15

Approach: Total = 10C2 = 45. Favourable = 4C2 = 6. P = 6/45 = 2/15.

Two dice are thrown together. What is the probability that the sum of numbers on the two faces is divisible by 4?

1/4

Approach: Divisible by 4: sums 4 (3 ways), 8 (5 ways), 12 (1 way) = 9 favourable. Total = 36. P = 9/36 = 1/4.

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